# chapter9 - Chapter 9 Solutions to Β§9.1 1 F(s = β β e...

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Unformatted text preview: Chapter 9 Solutions to Β§9.1 1. F(s) = β« β e βst e 2t dt = lim n ββ β« n e (2βs)t dt = 1 2 β s lim n ββ [ ] e (2βs)t n = 1 s β 2 if s > 2. 3. F(s) = β« β e βst sin btdt = lim n ββ β« n e βst sin btdt = lim n ββ e βst [βsin bt β bcos bt] s 2 + b 2 n = b s 2 + b 2 if s > 0. 5. F(s) = β« β e βst cosh btdt = lim n ββ β‘ β  n e βst e bt + e βbt 2 dt = lim n ββ e (b β s)t 2(b β s) β e β(b + s)t 2(b + s) n = lim n ββ e (b β s)n β 1 2(b β s) β e β(b + s)n + 1 2(b + s) = s s 2 β b 2 where s > |b|. 7. F(s) = β« β e βst (2t)dt = 2lim n ββ β« n e βst tdt = 2lim n ββ e βst (βst β 1) s 2 n = 2 lim n ββ βn se sn β 1 s 2 e sn + 1 s 2 = 2 s 2 where s > 0. 9. F(s) = β« 2 e βst (1)dt + β« 2 β e βst (β1)dt = e βst βs 2 + lim n ββ β« 2 n (βe βst )dt = [ e β2s βs + 1 s ] + lim n ββ e βst s 2 n = 1 s β 1 e 2s s + lim n ββ [ 1 se sn β 1 se 2s ] = 1 s (1 β 2e β2s ) provided s > 0. 11. F(s) = β« β e βst (e t sin t)dt = lim n ββ β« n e (1 β s)t sin tdt = lim n ββ e (1 β s)t [sin t β cos t] (s β 1) 2 + 1 n = lim n ββ e (1 β 2)n [sin n β cos n] (s β 1) 2 + 1 + 1 (s β 1) 2 + 1 = 1 (s β 1) 2 + 1 where s > 1. 267 268 Ch. 9 The Laplace Transform and Some Elementary Applications 13. L[2t β e 3t ] = L[2t] β L[e 3t ] = 2L[t] β 1 s β 3 = 2( 1 s 2 ) β 1 s β 3 = 2 s 2 β 1 s β 3 , where s > 3. 15. L[cosh bt] = L e bt 2 + e βbt 2 = 1 2 L[e bt ] + 1 2 L[e βbt ] = 1 2 1 s β b + 1 2 1 s + b = s s 2 β b 2 where s > |b|. 17. L[3t 2 β 5cos 2t + sin 3t] = 3L[t 2 ] β 5L[cos 2t] + L[sin 3t] = 3 2! s 3 β 5 s s 2 + 4 + 3 s 2 + 9 = 6 s 3 β 5s s 2 + 4 + 3 s 2 + 9 . 19. L[2e β3t + 4e t β 5sin t] = 2L[e β3t ] + 4L[e t ] β 5L[sin t] = 2 s + 3 + 4 s β 1 β 5 s 2 + 1 where s > 1. 21. L[4cos 2 bt] = 2L[2cos 2 bt] = 2L[cos 2bt + 1] = 2L[cos 2bt] + 2L[1] = 2 ( s 4b 2 + s 2 ) + 2 ( 1 s ) = 4(s 2 + 2b 2 ) (s 2 + 4b 2 )s where s > 0. 23. The graph of f is pieceβwise continuous on [0, β ). 25. The graph of f is not pieceβwise continuous on [0, β ), because lim n β1 + f(t) = β , i.e. the rightβ hand limit as t approaches 1 from above is not finite. 27. The graph of f is pieceβwise continuous on [0, β )....
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chapter9 - Chapter 9 Solutions to Β§9.1 1 F(s = β β e...

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